AP Physics C Mechanics Unit 5: Torque and Rotational Dynamics

Model rotation with angular kinematics, torque, rotational inertia, rotational equilibrium, and Newton’s laws in rotational form.

Covered syllabus points

5.1 Rotational Kinematics 5.2 Connecting Linear and Rotational Motion 5.3 Torque 5.4 Rotational Inertia 5.5 Rotational Equilibrium and Newton’s First Law in Rotational Form 5.6 Newton’s Second Law in Rotational Form

[Maximum number: 14]

A wind turbine includes a three-blade system that rotates about an axis through the end of each blade, as shown in Figure 1. Each blade has a length L and mass M, with a center of mass located at a distance $\frac{L}{3}$ from the axis of rotation, as shown in Figure 2.

(a)

Derive an expression for the rotational inertia of the three-blade system. Express your answer in terms of M, L, and physical constants, as appropriate. The rotational inertia of each blade about an axis through its center of mass is given by the equation $I_{\mathrm{cm}}=\frac{1}{18} M L^{2}$ .

[ 3 ]

For stating the parallel axis theorem

1 point

Example Response

For using correct substitutions of the rotational inertia of one blade about its center of mass and substituting the distance from the center of mass

1 point

Example Response

For multiplying the rotational inertia of one blade by 3

1 point

Example Response

$I_{\text {rotor }}=3 I_{\text {blade }}=\frac{1}{2} M L^{2}$

Example Solution

$I_{\text {blade }}=I_{\mathrm{CM}}+M d^{2}$ $I_{\text {blade }}=\frac{1}{18} M L^{2}+M\left(\frac{L}{3}\right)^{2}$ $I_{\text {blade }}=\frac{1}{6} M L^{2}$ $I_{\text {rotor }}=3 I_{\text {blade }}=\frac{1}{2} M L^{2}$

Total for part (a)

3 points

(b)

When the wind stops blowing, the angular speed of the system decreases. The angular speed $\omega$ of the system while slowing down is given as a function of time t by the equation $\omega=\omega_{0} e^{-\beta_{0} t}$ , where $\beta_{0}$ is a constant with appropriate units, as shown on the graph in Figure 3.

Figure 3

i. Calculate the amount of energy dissipated from t=0, when the wind stops blowing, until the system comes to rest.

ii. Derive an expression for the net torque exerted on the system as a function of t as the system slows down. Express your answers in terms of $\beta_{0}, \omega_{0}, M, L, I_{\text {sys }}$ , and physical constants, as appropriate.
iii. Derive an expression for the angular displacement of the system $\Delta \theta$ as a function of t. Express your answer in terms of $\beta_{0}, \omega_{0}, M, L, I_{\text {sys }}$ , and physical constants, as appropriate.

The three-blade system is now replaced with a second three-blade system identical to the first, except that the second three-blade system slows down according to the equation $\omega=\omega_{0} e^{-\beta t}$ , where $\omega_{0}=2.6 \mathrm{rad} / \mathrm{s}$ and $\beta>\beta_{0}$ . The original angular speed function is shown as a dashed line in Figure 4.

Figure 4

[ 9 ]

(i) The initial rotational kinetic energy is dissipated:
$E_{\mathrm{dis}}=-\frac12 I_{\mathrm{sys}}\omega_0^2 =-\frac12(6.7\times10^6)(2.6)^2 =-2.3\times10^7\ \mathrm{J}.$
(The dissipated-energy amount is (2.3\times10^7\ \mathrm J).)

(ii) Using $\tau=I_{\mathrm{sys}}\alpha$ ,
$\tau(t)=I_{\mathrm{sys}}\frac{d\omega}{dt} =-\beta_0 I_{\mathrm{sys}}\omega_0e^{-\beta_0t}.$

(iii)
$\begin{aligned} \Delta\theta(t)&=\int_0^t\omega_0e^{-\beta_0u}\,du\\ &=\frac{\omega_0}{\beta_0}\left(1-e^{-\beta_0t}\right). \end{aligned}$

(c)

On the graph in Figure 4, sketch the angular speed of the second three-blade system as a function of time t.

[ 2 ]

For drawing a continuous curve showing an exponential decay

1 point

For starting at $\omega=2.6 \mathrm{rad} / \mathrm{s}$ and drawing a curve below the original curve

1 point

Example Solution

Figure for Answer to Question (c) — AP Physics C: Mechanics

Total for part (d) for question 315 points

[Maximum number: 11]

Figure for Question 5 Torque and Rotational Dynamics question 2 — AP Physics C: Mechanics

Mech. 3. A uniform, rigid, thin board that is 2.4 m long and weighs 200 N is attached to a post by a pivot at point P and hangs over the edge of a building, as shown above. A crate weighing 400 N is attached to the right end of the board. The left end of the board, 0.80 m from the pivot, is attached to a light vertical wire anchored at the other end to keep the board from rotating.

(a)

Figure for Question (a) — AP Physics C: Mechanics

(b) Calculate the magnitudes of the forces exerted on the board by the post and by the wire. If you need to draw anything other than what you have shown in part (a) to assist in your solution, use the space below. Do NOT add anything to the figure in part (a).

[ 4 ]

(b)

Figure for Question (b) — AP Physics C: Mechanics

(c) The rotational inertia of the board alone about its center is $\frac{1}{12} M L^{2}$ , where M is the mass of the board and L is its length. Calculate the rotational inertia of the combined board-crate system about point P.

[ 3 ]

(c) 3 points For correctly calculating the rotational inertia of the board about P 1 point ( or $12.8 \mathrm{~kg} \cdot \mathrm{~m}^{2}$ using $g=10 \mathrm{~m} / \mathrm{s}^{2}$ ) For correctly calculating the rotational inertia of the crate about P 1 point (or $102.4 \mathrm{~kg} \cdot \mathrm{~m}^{2}$ using $g=10 \mathrm{~m} / \mathrm{s}^{2}$ ) For recognizing that the total rotational inertia of the system about P is the sum of the inertia of the board and the crate about P $\Sigma I_{P}=117.6 \mathrm{~kg} \cdot \mathrm{~m}^{2}$ (or $115.2 \mathrm{~kg} \cdot \mathrm{~m}^{2}$ using $g=10 \mathrm{~m} / \mathrm{s}^{2}$ )

(c)

Figure for Question (c) — AP Physics C: Mechanics

(d) Suppose that the wire breaks and the board begins to pivot about point P. Calculate each of the following.

[ 4 ]

(i)

Figure for Question (i) — AP Physics C: Mechanics

i. The magnitude of the initial angular acceleration of the board-crate system

[ 2 ]

i. 2 points For summing torques about P, setting the sum equal to $I \alpha$ , and substituting appropriate values $\Sigma \tau_{P}=w_{\text {board }} d_{\text {board }}+w_{\text {crate }} d_{\text {crate }}=I \alpha$ $(200 \mathrm{~N})(0.4 \mathrm{~m})+(400 \mathrm{~N})(1.6 \mathrm{~m})=115.2 \alpha$ For an answer consistent with part (c) $\alpha=6.13 \mathrm{rad} / \mathrm{s}^{2}$ (or $6.25 \mathrm{rad} / \mathrm{s}^{2}$ using $g=10 \mathrm{~m} / \mathrm{s}^{2}$ )

(ii)

Figure for Question (ii) — AP Physics C: Mechanics

ii. The magnitude of the initial linear acceleration of the left end of the board

[ 2 ]

ii. 2 points For a correct relationship between angular and linear acceleration $a=r \alpha$ $a=(0.80 \mathrm{~m})\left(6.13 \mathrm{rad} / \mathrm{s}^{2}\right)$ For an answer consistent with part (d)i. $a=4.9 \mathrm{~m} / \mathrm{s}^{2} \quad\left(\right.$ or $5.0 \mathrm{~m} / \mathrm{s}^{2}$ using $\left.g=10 \mathrm{~m} / \mathrm{s}^{2}\right)$

[Maximum number: 16]

A solid uniform disk is supported by a vertical stand. The disk is able to rotate with negligible friction about an axle that passes through the center of the disk. The mass and radius of the disk are given by $M_{\mathrm{d}}$ and R, respectively. The rotational inertia of the disk is $I_{\mathrm{d}}=\frac{1}{2} M_{\mathrm{d}} R^{2}$ . A string of negligible mass is draped over the disk and attached to the top of the disk at point P. One end of the string is connected to an unstretched ideal spring of spring constant k, which is fixed to the ground as shown in Figure 1.
A block of mass $m_{\mathrm{B}}$ is then attached to the string on the right side of the disk. The block is slowly lowered until the spring-disk-block system reaches equilibrium, as shown in Figure 2. In this equilibrium position, the disk has rotated clockwise through a small angle $\theta$ .
Give all algebraic answers in terms of $M_{\mathrm{d}}, R, k, \theta$ , and physical constants, as appropriate.

(a)

Derive an expression for the mass $m_{\mathrm{B}}$ of the block.

[ 3 ]

For indicating that the sum of the torques on the disk equals zero 1 point

\begin{aligned} & \Sigma \tau_{\text {on disk }}=0 \\ & \tau_{\mathrm{g}}=\tau_{\mathrm{s}} \end{aligned}

OR
For indicating that the sum of the forces equals zero

\begin{aligned} & \Sigma F=0 \\ & F_{\mathrm{g}}=F_{\mathrm{s}} \end{aligned}

For correctly substituting the expressions for the forces 1 point

\begin{aligned} & F_{\mathrm{g}} R=F_{\mathrm{s}} R \\ & m_{\mathrm{B}} g R=k \Delta x R \\ & m_{\mathrm{B}} g=k \Delta x \end{aligned}

\begin{aligned} & F_{\mathrm{g}}=F_{\mathrm{s}} \\ & m_{B} g=k \Delta x \end{aligned}

For correctly substituting for $\Delta x$ 1 point

\begin{aligned} & m_{\mathrm{B}} g=k R \theta \\ & m_{\mathrm{B}}=\frac{k R \theta}{g} \end{aligned}

Total for part (a)

3 points

(b)

At time t=0, the string on the right side of the disk is cut and the block falls to the ground. On the circle below, which represents the disk, draw and label the forces (not components) that act on the disk immediately after the string is cut and the block is falling to the ground. Each force should be represented by an arrow that starts on and is directed away from the point of application.

[ 3 ]

For drawing and labeling the force of the tension exerted on the disk anywhere between point P and the left edge of the disk, including point P and the left edge of the disk, tangent to the disk

1 point

For correct location and label of the force due to gravity exerted on the disk, directed straight
down

1 point

For correct location and label of the force exerted on the disk by the axle, directed such that
the disk remains in translational equilibrium (i.e., $\Sigma F=0$ )

1 point

Example Response

Figure for Answer to Question (b) — AP Physics C: Mechanics

Scoring Notes:

- Examples of appropriate labels for the force due to gravity include: $F_{\mathrm{G}}, F_{\mathrm{g}}, F_{\text {grav }}$ , W, m g, M g, "grav force," "F Earth on block," "F on block by Earth," $F_{\text {Earth on block }}$ , $F_{\mathrm{E}, \text { Block }}$ . The labels G and g are not appropriate labels for the force due to gravity. $F_{\mathrm{n}}, F_{\mathrm{N}}, N$ , "normal force," "ground force," or similar labels may be used for the normal force, which can be used instead of $F_{\text {axle }} \cdot F_{\text {spring }}, F_{\mathrm{s}}, T_{\text {spring }}, T$ , "spring force," or similar labels may be used for the tension force exerted by the spring.

- A response with extraneous forces or vectors can earn a maximum of two points.

Total for part (b)

3 points

(c)

Derive an expression for the angular acceleration $\alpha$ of the disk immediately after the string is cut.

[ 5 ]

For indicating that the net torque is due only to the force exerted on the disk by the tension in the rotational form of Newton's second law
$\tau_{\mathrm{s}}=I_{\mathrm{d}} \alpha$

1 point

For correctly expressing the torque on the disk by the tension in terms of the spring force, which is equal to the tension, and the lever (moment) arm
$F_{\mathrm{s}} R=I_{\mathrm{d}} \alpha$

1 point

For correctly substituting for $F_{s}$ $-k \Delta x R=I_{\mathrm{d}} \alpha$

1 point

For correctly substituting $I_{\mathrm{d}}$ and $\Delta x$ , or an expression for $\Delta x$ consistent with part (a)
$\begin{aligned} -k(R \theta) R=\frac{1}{2} M_{\mathrm{d}} R^{2} \alpha \alpha=-\frac{2 k \theta}{M_{\mathrm{d}}} \end{aligned}$

1 point

Scoring Note: The negative sign is not necessary to earn this point.

(d)

At $t=t_{1}$ , the disk has rotated and point P is again directly above the axle. Sketch a graph of the magnitude of the angular velocity $\omega$ of the disk as a function of time t from t=0 to $t=t_{1}$ .

Figure for Question (d) — AP Physics C: Mechanics

Figure 3

Note: Figure not drawn to scale.

[ 2 ]

For a sketch that starts at zero and monotonically increases until time $t=t_{1}$ 1 point
For a sketch that is concave down between time t=0 and $t=t_{1}$ 1 point

Example Response

Figure for Answer to Question (d) — AP Physics C: Mechanics

Scoring Note: Any part of the graph beyond $t_{1}$ is not considered in scoring.

Total for part (d)

2 points

(e)

The disk is adjusted on the support so that the axle does not pass through the center of mass of the disk. The block is again hung on the right side of the disk and the spring-disk-block system comes to equilibrium, as shown in Figure 3. The axle does not exert a torque on the disk. For each force on the disk, indicate whether the magnitude of the torque about the axle caused by that force increases, decreases, or stays the same relative to part (b).

[ 3 ]

[Maximum number: 11]

A uniform rod of length L and mass m is attached to a pivot on a vertical pole, as shown in Figure 1. There is negligible friction between the rod and the pivot. A horizontal string connects Point Q on the rod to the pole. The rod makes an angle $\theta$ with the pole. A block of mass 3 m hangs from the rod at Point P. The center of mass of the rod is located at Point C.

(a)

In Figure 1, Point P is located $\frac{3}{8} L$ from the pivot and Point Q is located $\frac{6}{8} L$ from the pivot. Derive an equation for the tension $F_{\mathrm{T}}$ in the horizontal string in terms of $L, m, \theta$ , and physical constants, as appropriate.

Figure 2
Note: Figure not drawn to scale.

[ 3 ]

For indicating that the net torque exerted on the rod is equal to zero 1 point
Example Response

\Sigma \tau=0

For a correct expression for the torque exerted on the rod by the hanging mass 1 point
Example Response

3 m g \sin \theta\left(\frac{3 L}{8}\right)

For a correct expression for the torque exerted on the rod by the gravitational force 1 point
Example Response

m g \sin \theta\left(\frac{4 L}{8}\right)

For a correct expression for the torque exerted on the rod by the string 1 point
Example Responses

F_{\mathrm{T}} \sin \left(90^{\circ}-\theta\right)\left(\frac{6 L}{8}\right) \quad \text { OR } \quad F_{\mathrm{T}} \cos \theta\left(\frac{6 L}{8}\right)

Example Solution

\begin{aligned} & \Sigma \tau=0 \\ & 3 m g \sin \theta\left(\frac{3 L}{8}\right)+m g \sin \theta\left(\frac{4 L}{8}\right)-F_{\mathrm{T}} \sin \left(90^{\circ}-\theta\right)\left(\frac{6 L}{8}\right)=0 \\ & 3 m g \sin \theta\left(\frac{3 L}{8}\right)+m g \sin \theta\left(\frac{4 L}{8}\right)-F_{\mathrm{T}} \cos \theta\left(\frac{6 L}{8}\right)=0 \\ & \left(\frac{9}{8}\right) m g \sin \theta+\left(\frac{4}{8}\right) m g \sin \theta=\left(\frac{6}{8}\right) F_{\mathrm{T}} \cos \theta \\ & 13 m g \sin \theta=6 F_{\mathrm{T}} \cos \theta \\ & F_{\mathrm{T}}=\frac{13}{6} m g \tan \theta \end{aligned}

Scoring Note: A maximum of three points may be earned if the trigonometric functions ( $\sin$ and $\cos$ ) are reversed for all three torque terms.

(b)

The original string is replaced with a longer string that connects Point Q to a higher location on the vertical pole, as shown in Figure 2. The angle $\theta$ remains the same. How does the new tension $F_{\mathrm{T}, \text { new }}$ compare with the original tension $F_{\mathrm{T}}$ from part (b) ? Justify your reasoning.

Figure 3 Note: Figure not drawn to scale.

[ 2 ]

For indicating that the torque exerted on the rod by the string is always the same

1 point

For stating that as the angle between the string and the rod increases, the force exerted on the rod by the string decreases

1 point

Example Response

Because the torque exerted on the rod by the string is always the same, as the angle between the string and the rod increases, the tension $F_{T, \text { new }}$ must decrease.

Total for part (c)

2 points

(c)

A nonuniform rod is now attached to the pivot, as shown in Figure 3. There is negligible friction between the nonuniform rod and the pivot. The rod has a length of 1.2 m and a linear mass density $\lambda(x)=A+B x$ , where x is the distance from the pivot, $A=6.0 \mathrm{~kg} / \mathrm{m}$ , and $B=10.0 \mathrm{~kg} / \mathrm{m}^{2}$ .

[ 6 ]

(i)

Calculate the mass of the rod.

[ 3 ]

For indicating the total mass is the sum of differentiable masses along the length of the rod

1 point

Example Response
$M=\int d m$

For correctly writing d m in terms of x

1 point

Example Response
$M=\int_{0}^{1.2}(6+10 x) d x$

For a correct numeric answer with correct units

1 point

Example Response
$M=14.4 \mathrm{~kg}$

Example Solution
$\begin{aligned} M=\int d m M=\int \lambda d x M=\int_{0}^{1.2}(6+10 x) d x M=\left.\left(6 x+\frac{10 x^{2}}{2}\right)\right|_{0} ^{1.2 \mathrm{~m}} M=6 \mathrm{~kg} / \mathrm{m}(1.2 \mathrm{~m})+\frac{10 \mathrm{~kg} / \mathrm{m}^{2}(1.2 \mathrm{~m})^{2}}{2} M=14.4 \mathrm{~kg} \end{aligned}$

(ii)

Calculate the rotational inertia of the rod about the pivot.

[ 3 ]

(ii) For a correct substitution of $\lambda$ into an integral expression of rotational inertia 1 point
Example Response

I=\int_{0}^{1.2 \mathrm{~m}}(A+B x) x^{2} d x

For a correct integration 1 point
Example Response

I=\left.\left(\frac{A x^{3}}{3}+\frac{B x^{4}}{4}\right)\right|_{0} ^{1.2 \mathrm{~m}}

For a correct numeric answer with correct units 1 point
Example Response

I=8.64 \mathrm{~kg} \cdot \mathrm{~m}^{2}

Example Solution

\begin{aligned} & I=\int r^{2} d m \quad d m=\lambda d r \text { and } r=x \\ & I=\int_{0}^{1.2 \mathrm{~m}} \lambda x^{2} d x \\ & I=\int_{0}^{1.2 \mathrm{~m}}(A+B x) x^{2} d x \\ & I=\left.\left(\frac{A x^{3}}{3}+\frac{B x^{4}}{4}\right)\right|_{0} ^{1.2 \mathrm{~m}} \\ & I=\frac{(6.0 \mathrm{~kg} / \mathrm{m})(1.2 \mathrm{~m})^{3}}{3}+\frac{\left(10.0 \mathrm{~kg} / \mathrm{m}^{2}\right)(1.2 \mathrm{~m})^{4}}{4} \\ & I=8.64 \mathrm{~kg} \cdot \mathrm{~m}^{2} \end{aligned}

Total for part (d) for question 315 points

All question bank results loaded

Covered syllabus points

5 Torque and Rotational Dynamics question 1

Question (a)

Question (b)

Question (c)

5 Torque and Rotational Dynamics question 2

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

5 Torque and Rotational Dynamics question 3

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

5 Torque and Rotational Dynamics question 4

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)